A Cessna airplane can accelerate from rest to a ground speed of 78 knots in 29.8 s. What is the average acceleration of this airplane?
I figured out that 78 knots is equal to 89.76 miles, 144.45 km, or 144,450 m. And I think the equation might be average=delta V over delta t. But I'm not sure if this is right...help!
You are thinking correctly. Delta v over delta t expresses acceleration.
delta v = 78kts (nautical miles/second) (convert to other units if desired.
delta t = 29.8s
average acceleration = (78/29.8) nautical miles/sec^2.
I try this and I get 2.62 knots/s^2, but it says this answer is wrong. Is it?
N/m, I needed to convert to knots to meters per second. I got the right answer.
Thanks for the help, Quidditch. :)
To find the average acceleration of the airplane, you are correct in using the equation average acceleration (a) = change in velocity (ΔV) / change in time (Δt).
In this case, the change in velocity (ΔV) is the difference in the ground speed of the airplane. Starting from rest (0 knots), the airplane reaches a ground speed of 78 knots, so ΔV = 78 knots - 0 knots = 78 knots.
The change in time (Δt) is given as 29.8 seconds.
Using the equation average acceleration (a) = ΔV / Δt, we can substitute the values:
average acceleration (a) = 78 knots / 29.8 s
Now, we need to convert the units of knots to m/s to ensure consistent units. 1 knot is approximately equal to 0.5144 m/s.
average acceleration (a) = (78 knots * 0.5144 m/s) / 29.8 s
Simplifying the equation:
average acceleration (a) = 1.0 m/s^2 (approximately)
Therefore, the average acceleration of the Cessna airplane is approximately 1.0 m/s².